boolean-algebra

Boolean algebra simplification exercise

So I have been given the following expression, but I cannot seem to solve it, can anyone do this and show the steps please?

Prove XY'Z + XYZ' + XYZ = XY + XZ

Solution

  • XY'Z + XYZ' + XYZ = XY + XZ

    1. Notice X and Z are common factors between XY'Z and XYZ.

    XZ(Y' + Y) + XYZ' =

    1. Y' + Y is equal to 1 (if Y=0 then Y'=1 and so 0 + 1 = 1, that is 0 or 1 = 1. Similarly, if Y=1 then Y'=0 and so 1 + 0 = 1). Therefore, what you get is:

    XZ·1 + XYZ' =

    1. XZ·1 = XZ since A·1 = A (if A=0 then 0·1 is 0 and if A=1 then 1·1 = 1). Now the function is simplified to:

    XZ + XYZ' =

    1. Notice once again X is a common factor between XZ and XYZ'.

    X(Z + YZ') =

    1. Notice this time that Z + YZ' is a special case of the distributive law, which is A + A'B = A + B. This is because if we apply the general distributive law A + BC = (A + B)·(A + C) then we get A + A'B = (A + A')·(A + B) = 1·(A + B) = A + B. Following this reasoning we get to simplify the function even further:

    X(Z + Y) =

    1. All that's left is for us to use the distributive law and we finally arrive to the final result:

    XY + XZ

    Please note that nothing is written between variables, an AND operator (or "·" symbol) is assumed. It's just a way to save space.